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Five-digit numbers are formed from digits 1 to 9, with no digit repeat

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sherxon
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Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   13 Jun 2021, 23:03
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Five-digit numbers are formed from digits 1 to 9, with no digit repeated more than once. How many numbers contain 2, 4 and 5 at the same time?

A. 900
B. 1200
C. 1500
D. 1800
E. 2100
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D
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chetan2u
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Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   13 Jun 2021, 23:32
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sherxon wrote:
Five-digit numbers are formed from digits 1 to 9, with no digit repeated more than once. How many numbers contain 2, 4 and 5 at the same time?

A. 900
B. 1200
C. 1500
D. 1800
E. 2100


Thus, we have to chose 2 digits from remaining 6 digits, as 2, 4 and 5 are already picked. => 6C2=15

As we are looking at arrangements of these chosen 5 digits, it can be done in 5! ways.

Total = 15*5!=15*120=1800

D
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SatvikVedala
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Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   05 Jan 2023, 09:42
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sherxon wrote:
Five-digit numbers are formed from digits 1 to 9, with no digit repeated more than once. How many numbers contain 2, 4 and 5 at the same time?

A. 900
B. 1200
C. 1500
D. 1800
E. 2100


_ _ _ _ _ (five digit number)

information - 1 to 9 digits only & each number can be used once

Condition - numbers must contain 2,4 & 5 at the same time

resolving condition first

a) considering 2 occupies first - 2 has 5 places to occupy
b) considering 4 occupies next - 4 has 4 places to occupy
c) considering 5 occupies next - 5 has 3 places to occupy

Therefore no. of ways 2,4 & 5 can be at the same time ---> 5*4*3 = 60 ways

From information - we have 6 other numbers (1,3,6,7,8,9) & 2 Places

Therefore no. of ways to occupy 2 places ---> 6*5 = 30 ways

Total ways = 30*60 = 1800 Ways
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Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   14 Jun 2021, 01:21
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Solution:

We need to make 5 digit number from digits 1 to 9 which will definitely have 2, 4 and 5.
3 out of 5 digit are already selected.
So we need to choose the rest 2 digit from 6 remaining digits. Which can be done in = \(6C2\) ways.

Now after choosing we also need to rearrange them in al possible ways. So 5 digits can be rearranged in = \(5!\) ways.

Thus the final answer \(= 6C2 * 5! = 15 * 120 = 1800\).

Hence the right answer is Option D.
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Surgicalgoblin
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Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   17 Jun 2021, 20:23
I have a doubt, all the numbers can be repeated once means that we can reuse the numbers just once or no repetition is allowed and if so how does each option affect the answer ?
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Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   05 Sep 2021, 22:34
is there any other way solve this question?
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summerbummer
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Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink] New post   12 Sep 2021, 02:32
sherxon wrote:
Five-digit numbers are formed from digits 1 to 9, with no digit repeated more than once. How many numbers contain 2, 4 and 5 at the same time?

A. 900
B. 1200
C. 1500
D. 1800
E. 2100

---------------------------------------------

Can someone show using filling the blanks method?
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gmatclubot
Re: Five-digit numbers are formed from digits 1 to 9, with no digit repeat [#permalink]
12 Sep 2021, 02:32
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